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Knowledge and practice are one. ..."Wang Yang Ming (1472-1529, Chinese Philosopher) "
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Author Topic: Conservation of Angular momentum and 3D circular motion  (Read 15316 times)
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Fu-Kwun Hwang
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on: March 12, 2010, 11:06:08 pm » posted from:Taipei,T\'ai-pei,Taiwan

A particle with mass m is moving with constant speed v along a circular orbit (radius r).
The centripetal force F=m\frac{v^2}{r} is provided by gravitation force from another mass M=F/g.
A string is connected from mass m to the origin then connected to mass M.
Because the force is always in the \hat{r} direction, so the angular momentum \vec{L}=m\,\vec{r}\times \vec{v} is conserved. i.e. L=mr^2\omega is a constant.
 
For particle with mass m:

 m \frac{d^2r}{dt^2}=m\frac{dv}{dt}= m \frac{v^2}{r}-Mg=\frac{L^2}{mr^3}- Mg
 \omega=\frac{L}{mr^2}

The following is a simulation of the above model.

You can change the mass M or the radius r with sliders.
The mass M also changed to keep the mass m in circular motion when you change r.
However, if you change mass M , the equilibrium condition will be broken.

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Reply #1 on: March 19, 2010, 10:53:15 am » posted from:Yogyakarta,Yogyakarta,Indonesia

there is more simple formula for this simulation Huh
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Reply #2 on: March 19, 2010, 01:22:45 pm » posted from:Taipei,T\'ai-pei,Taiwan

Quote
there is more simple formula for this simulation

Could you explain what do you mean in detail?
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Reply #3 on: July 07, 2010, 12:32:23 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE

question:
how is the tension in the string calculated?

My understanding:
T = m*v*v/r; // added lookang for tension only work for circular motion

my hypothesis:
T1 = m*v*v/r + m*dvr/dt ? // radial acceleration due to tangential change in velocity and radial acceleration due to radial change in velocity ?

where dvr/dt = (cst*cst/(m*r*r*r)-M*g)/m

My equation:
T1 = m*v*v/r +(cst*cst/(m*r*r*r)-M*g) ;
can help take a look if my equation of Tension is correct ?
did i get the generalized equation for tension in any elliptical motion?
the logic seems correct and i implement the equation looks correct. is it correct conceptually in the absence of frictional forces Smiley


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« Last Edit: July 07, 2010, 12:54:54 pm by lookang » Logged
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Reply #4 on: July 07, 2010, 06:05:49 pm » posted from:Taipei,T\'ai-pei,Taiwan

If the mass is in circular motion, then the tension T=m\frac{v^2}{r}.

What if we increase the tension so that T>m\frac{v^2}{r}
Will the mass moving inward(\frac{dv_r}{dt}<0) or moving outward(\frac{dv_r}{dt}>0)?

Thank about it and then check with your equation!
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Reply #5 on: July 07, 2010, 06:16:08 pm »

If the mass is in circular motion, then the tension T=m\frac{v^2}{r}.

What if we increase the tension so that T>m\frac{v^2}{r}.
Will the mass moving inward(\frac{dv_r}{dt}<0) or moving outward(\frac{dv_r}{dt}>0)?

Thank about it and then check with your equation!
I see I got the sign convention wrong!
I forgot the convention for r unit vector is positive outward.
Will change it to
 T1 = m*v*v/r -(cst*cst/(m*r*r*r)-M*g) ;
thanks !!! Smiley

your masterful questioning allows to think and figure out the answer myself.
Appreciate your superb advise Smiley

my remixed applet is here
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0
enjoy!
« Last Edit: July 07, 2010, 06:21:45 pm by lookang » Logged
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Knowledge and practice are one. ..."Wang Yang Ming (1472-1529, Chinese Philosopher) "
 
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